We are given two piecewise functions:

1. ( f(x) = \begin{cases} x + 5 & \text{if } x < -2 \ x^2 + 2x + 3 & \text{if } x \geq -2 \end{cases} )

2. ( g(x) = \begin{cases} -6 & \text{if } x < -2 \ (x + 1)^4 & \text{if } x \geq -2 \end{cases} )

We need to calculate $f(3) + g(-2)$.

Step 1: Calculate $f(3)$

Since $3 \geq -2$, we use the second piece of $f(x)$: $f(3) = 3^2 + 2(3) + 3 = 9 + 6 + 3 = 18$

Step 2: Calculate $g(-2)$

Since $-2 \geq -2$, we use the second piece of $g(x)$: $g(-2) = (-2 + 1)^4 = (-1)^4 = 1$

Step 3: Add the results

$f(3) + g(-2) = 18 + 1 = 19$

Thus, $f(3) + g(-2) = 19$.

Would you like more details or have any questions about this solution?

Here are 5 related questions you can explore:

1. What happens if $x = -2$ in both $f(x)$ and $g(x)$?
2. How would you find the derivative of the piecewise function $f(x)$?
3. What is the value of $f(-3) + g(-3)$?
4. Can you graph both $f(x)$ and $g(x)$ to explore their behavior visually?
5. How would you evaluate $f'(x)$ and $g'(x)$ at specific points?

Tip: Always check which condition applies in piecewise functions before solving!